## 國 立 交 通 大 學

### 應 用 數 學 系

### 碩 士 論 文

### Vogan 圖之距離

### The Distance of Vogan Diagram

### 研 究 生： 程千砡

### 指導教授： 蔡孟傑 教授

### Vogan 圖之距離

### The Distance of Vogan Diagram

### 研 究 生：程千砡 Student：Chien-Yu Chen

### 指導教授：蔡孟傑教授 Advisor：Dr. Meng-Kiat Chuah

### 國 立 交 通 大 學

### 應 用 數 學 系

### 碩 士 論 文

### A Thesis

### Submitted to Department of Applied Mathematics

### College of Science

### National Chiao Tung University

### In partical Fulfillment of Requirement

### For the Degree of Master

### In

### Applied Mathematics

### June 2006

### Vogan 圖之距離

### 研究生：程千砡 指導教授：蔡孟傑 教授

### 國 立 交 通 大 學

### 應 用 數 學 系

### 摘 要

### Vogan 圖為一個含有 involution 的 Dynkin 圖，並且被固定住的頂點可以是黑

色的。若一個 Vogan 圖可以被表示成另一個 Vogan 圖，則他們可以說是等價的。任何 一種含有多個黑色頂點的 Vogan 圖都會與某個 Vogan 圖等價，而這個 Vogan 圖只有一 個黑色頂點。我們的目的便是要找出含有多個黑色頂點的 Vogan 圖需要多少步驟，才 能找到與他等價的單黑色頂點之 Vogan 圖。 這篇論文分為兩個部份，在第一部份(第一節至第五節)，我們會對 Vogan 圖先做 一些簡單的介紹以及 Vogan 圖距離的定義，再來便會以 Vogan 圖中幾個典型的圖來證 明他們的距離問題；最後的部份(第六節至第九節)則是證明 Vogan 圖中其他特別的圖 之距離。### 中華民國九十五年六月

### The Distance of Vogan Diagrams

### Student: Chien-Yu Chen Advisor: Dr. Meng-Kiat Chuah

### Department of Applied Mathematics

### National Chiao Tung University

### Abstract

A Vogan diagram is a Dynkin diagram with an involution, andthe vertices fixed by the involution may be black. If a Vogandiagram can be represented by another Vogan diagrams, then they are equivalent. Any Vogan diagram with many black vertices is equivalent to a diagram with only one black vertex. Our purpose is to find the steps from a Vogan diagram with many black vertices to one black vertex.

This thesis is divided into two parts. In the first part, consisting of Sections 1-5, we give a brief introduction of some fundamental concepts in Vogan diagram and the distance of two Vogan diagrams in classical types with proof. In the last part, Sections 6-9, we prove the distance of two Vogan diagrams in exceptional types.

### 誌 謝

兩年的研究生生活，稍縱即逝，很感謝當初蔡孟傑教授願意當我的指導教授，並 且悉心教導，讓我能順利完成論文，除此之外，老師也會帶我們去打打羽球，培養感 情之餘順便磨練球技。感謝博士班的舉卿學姊，當課業上遇到問題，不論時間早晚， 她總是很有耐心教我，她也會告訴我哪裡有好玩好吃的，是位個性開朗、很照顧人的 學姐。 我很幸運有一群貼心的同學和室友，陪我分享一切，談談心事、談談未來，偶爾 也會一起打羽球，抒解身心，讓我的研究所生活一點也不孤單。我更要感激采瑩，她 從不吝於將自己所知道的教給我，也因為有她的鼓勵和打氣，讓我們能一塊完成論 文，一塊迎接口試過關的喜悅。謝謝凱奇學長教導我很多事情，除了論文，也給予我 精神上很大的支持。 求學路程一路走來，遇到再大的困難，都有家人在背後支撐著我，給我鼓勵，真 的很感謝媽媽、爸爸、哥哥、跟妹妹。我是個很戀家的人，即使家住南部仍常每個月 都會回家一次，媽媽為了不讓我坐車辛苦，偶爾會北上來看我，還帶很多好吃給我， 彼此聊著生活上的大小事，因為母親的相伴，讓我覺得讀書是件很幸福的事。 最後我將本論文的成果獻給我的父母、師長以及所有關心我的朋友，希望一起分 享完成論文的這份喜悅。**Contents **

### 1 Introduction 1

### 2 Type A 4

### 3 Type B 10

### 4 Type C 13

### 5 Type D

### 17

### 6 Type E

6### 26

### 7 Type E

7### 30

### 8 Type E

8### 32

### 9 Type F

4### and G

2### 34

### Reference

### 36

### 1

### Introduction

A Dynkin diagram is a certain type of graph. It satisfies the property that two
vertices may be connected by 0, 1, 2 or 3 edges, and an orientation (arrow) is assigned
to each double or triple-edge. There are altogether seven classes of Dynkin diagrams,
*labelled as types A, B, C, D, E, F, G ([2, Chapter 11]). A Dynkin diagram in each type*
is specified by a subscript which indicates the number of vertices in that diagram,
*for example A*5 *is the diagram of type A which has 5 vertices. Then all the Dynkin*

*diagrams are given by An, Bn, Cn, Dn* *for n ≥ 1, as well as E*6*, E*7*, E*8*, F*4*, G*2. The

*infinite lists of diagrams of types A, B, C, D are called classical, and the finite lists of*
*types E, F, G are called exceptional.*

A Vogan diagram is a Dynkin diagram together with extra information. Namely,
*there is a diagram involution θ, such that the fixed points of θ are colored white or*
*black [3]. Many Dynkin diagrams, for example Bn* *and Cn*, have trivial symmetry. In

these cases the involution is the identity.

This thesis studies the following algorithm on the Vogan diagrams. Suppose that

*v is a Vogan diagram, and p is a black vertex of v. Let Fp* be the algorithm which

*reverses the colors of all the θ-fixed vertices which are adjacent to p (but not p itself),*
*except when the vertex is joint to p with a double-edge with arrow pointing towards*

*p. In this way, Fp(v) is another Vogan diagram. We give some examples as follows.*

*Example. The vertices are labelled 1, 2, . . . starting from the left.*

*v =* f v v f
*⇒ F*2*(v) =* v v f f
*⇒ F*3*(v) =* f f v v
*u =* f v f v* _{=⇒}*v

*⇒ F*4

*(u) =*f v v v

*=⇒*f

*⇒ F*5

*(u) =*f v f v

*=⇒*v

algorithms

*v = v*0 *→ v*1 *→ ... → vk* *= w* (1.1)

*such that each step vi* *→ vi+1* *is given by some Fp. If v and w are equivalent and k in*

*(1.1) is as small as possible, we say that k = d(v, w) is the distance between v and w.*
Clearly a diagram without black vertex is not equivalent to any other one. Therefore,
once and for all, we may consider only Vogan diagrams with black vertices, and denote
*them by V (X), where X is a Dynkin diagram with trivial diagram involution. Let*

*V*1*(X) ⊂ V (X) denote the diagrams with exactly one black vertex. It is known that*

*every diagram in V (X) is equivalent to a diagram in V*1*(X) [1][3]. It allows us to*

*define the distance between v ∈ V (X) and V*1*(X) by*

*d(v, V*1*(X)) = min*

*w∈V*1*(X)*

*d(v, w).* (1.2)
*For fixed X, we intend to seek an upper bound for {d(v, V*1*(X)); v ∈ V (X)}, namely*

*d(V (X)) = max*

*v∈V (X)d(v, V*1*(X)).* (1.3)

We present the main results of the thesis as follows. The classical Vogan diagrams
*are denoted by V (An), V (Bn), V (Cn), V (Dn), V (Dn, θ), where θ denotes the nontrivial*

*involution. We need not consider V (An, θ) because it contains only one diagram.*

Theorem 1

*(a) d(An) is bounded above by (n − 1)n*2*.*

*(b) d(Bn) is bounded above by (n − 1)n*2*.*

*(c) d(Cn) is bounded above by (n − 1)(n − 1)*2*.*

*(d) d(Dn) is bounded above by (n − 1)(n − 2)*2*.*

*(e) d(Dn, θ) is bounded above by (n − 1)(n − 2)*2*.*

*For the proof of Theorem 1, we shall study type A in Section 2, type B in Section*
*3, type C in Section 4, and type D in Section 5.*

*For the exceptional diagrams, only E*6*contains nontrivial involution θ. We provide*

Theorem 2

*(a) d(E*6*) is bounded above by 130;*

*(b) d(E*6*, θ) is bounded above by 1;*

*(c) d(E*7*) is bounded above by 228;*

*(d) d(E*8*) is bounded above by 180;*

*(e) d(F*4*) is bounded above by 4;*

*(f ) d(G*2*) is bounded above by 1.*

*For the proof of Theorem 2, we shall study E*6 *in Section 6, E*7 *in Section 7, E*8

### 2

### Type A

*Recall that a Dynkin diagram of type An* is given by

h h p p p h h

1 2 *n − 1* *n*

(2.1)
*A Vogan diagram of An* is labelled as (2.1) with at least one black vertex, denoted

*by V (An*). According to the theorem by Borel and de Siebenthal[1, Theorem 6.96],

every Vogan diagram is equivalent to a Vogan diagram with 0 or 1 painted vertex. For
*type An*, all Vogan diagrams with some black vertices can be equivalent to a painted

*one. Let Vk(An) ⊂ V (An) denote the Vogan diagram of An* *with k black vertices.*

*From (1.2), it also extends to the distance between Vk(An) and V*1*(An*) by

*d(Vk(An), V*1*(An)) = min{d(v, w); v ∈ Vk(An), w ∈ V*1*(An)}* (2.2)

*or just write dk(V*1*(An*)).

*For convenience, we write (i*1*, . . . , ik) ∈ V (An*) to denote the diagram with vertices

*i*1*, . . . , ik* *painted. For example (3, 5) ∈ V (A*6) means f f v f v f

*In this section, we will discuss the distance between Vk(An) and V*1*(An*); in other

words, we provide an upper bound for it.

*d(V (An*)) = max
*v∈V (An*)

*d(v, V*1*(An)).* (2.3)

*Example. For V (A*4), there are 7 cases to discuss.

*(1, 2) → (1)*

*(1, 3) → (1, 2, 3) → (2)*

*(1, 4) → (1, 2, 4) → (2, 3, 4) → (3)*
*(2, 3) → (1, 2) → (1)*

*(1, 2, 4) → (2, 3, 4) → (3)*

*(1, 2, 3, 4) → (2, 4) → (2, 3, 4) → (3)*

*By (2.3), d(V (A*4*)) = max{1, 2, 3, 2, 1, 2, 3} = 3.*

*Proposition 2.1 Let u = (i, j) ∈ V (An*)

*(a) If i − 1 ≤ n − j, then d(u, v) ≤ i(j − i) where v = (j − i) ∈ V (An).*

*(b) If i−1 > n−j, then d(u, w) ≤ (n−j+1)(j −i) where w = (n+i−j +1) ∈ V (An).*

*P roof : u has two black vertices in V (An*), and it can be equivalent to a diagram in

*V*1*(An); in other words, u ∼ s ∈ V*1*(A).*

*There are two ways to make u reduce to a Vogan diagram of An* with only one

painted vertex:

*Case 1 Move i and j towards the left:*

e e p p p u p p p u p p p e
1 *i* *j* *n* *∼*
e e p p p u p p p u p p p e
1 *i − 1* *j − 1* *n*
*∼* e p p p u p p p u p p p p p p e
1 *i − 2* *j − 2* *n*
*∼* p p p p p p
*∼* u e p p p u e p p p e
1 *j − i + 1* *n*
*∼* p p p p p p
*∼* e p p p u e p p p p p e
*j − i*
1 *n*
*u = (i, j) −→ (i − 1, i, i + 1, j) −→ (i − 1, i + 1, i + 2, j) −→ · · · −→ (i − 1, j − 2,*
*j − 1, j) −→ (i − 1, j − 1) needs at most j − i steps.*

Continuously,

*(i − 1, j − 1) −→ (i − 2, i − 2) needs at most j − i steps.*
*(i − 2, j − 2) −→ (i − 3, i − 3) needs at most j − i steps.*

...

*(2, j − i + 2) −→ (1, j − i + 1) needs at most j − i steps.*
Finally,

*(1, j−i+1) −→ (1, 2, j−i+1) −→ (2, 3, j−i+1) −→ ··· −→ (j−i−1, j−i, j−i+1) −→*
*(j − i) needs at most j − i steps.*

*Thus, d(u, (j − i)) ≤ (j − i)[(i − 1) + 1] = i(j − i). Let v = (j − i) and d(u, v) = d*1.

*Case 2 Move i and j towards the right:*

e e p p p u p p p u p p p e
1 *i* *j* *n* *∼*
e p p p u p p p u p p p p p p e
1 *i + 1* *j + 1* *n*
*∼* e p p p p p p u p p p u p p e
1 *i + 2* *j + 2* *n*
*∼* p p p p p p
*∼* e e p p p u e p p p u
1 *n + i − j* *n*
*∼* p p p p p p
*∼* e p p p p p p e u p p p e
*n + i − j + 1*
1 *n*

*u = (i, j) −→ (i, j − 1, j, j + 1) −→ (i, j − 2, j − 1, j + 1) −→ · · · → (i, i + 1, i + 2, j + 1)*
*−→ (i + 1, j + 1) needs at most j − i steps.*

*(i + 1, j + 1) −→ · · · −→ (i + 2, j + 2) needs at most j − i steps.*
*(i + 2, j + 2) −→ · · · −→ (i + 3, j + 3) needs at most j − i steps.*

...

*(n + i − j − 1, n − 1) −→ (n + i − j, n) needs at most j − i steps.*
Finally,

*(n + i − j, n) −→ (n + i − j, n − 1, n) −→ (n + i − j, n − 2, n − 1) −→ · · · −→*
*(n + i − j, n + i − j + 1, n + i − j + 2) −→ (n + i − j + 1) needs at most j − i steps.*
*Thus,d(u, (n + i − j + 1)) ≤ (j − i)[(n + i − j + i) + 1] = (n − j + 1)(j − i). Let*

*w = (n + i − j + 1) and d(u, w) = d*2.

Comparison with Case 1 and Case 2, the proof can be completed as the following:
*(a) i − 1 ≤ n − j ⇒ i < n − j + 1⇒ i(j − i) ≤ (n − j + 1)(j − i) ⇒ d*1 *≤ d*2 with

*j − i > 0. By (1.2), it needs at most i(j − i) steps s.t. u → · · · → v = (j − i) ∈ V*1*(An*).

*(b) i − 1 > n − j ⇒ i > n − j + 1⇒ i(j − i) > (n − j + 1)(j − i) ⇒ d*1 *≤ d*2 with

*j − i > 0. By (1.2), it needs at most (n − j + 1)(j − i) steps such that u → · · · →*
*v = (j − i) ∈ V*1*(An*).

Therefore, we can conclude that:

*Proposition 2.2 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(An), then dk(V*1*(An)) ≤ (k − 1)n*2*.*

*P roof : By Proposition 2.1, we have known that any two painted vertices which can*

*be reduced to be only one painted vertex in Vk(An) needs at most d*1 *or d*2 steps.

*Case 1 If i*1 *− 1 ≤ n − ik, and just move the leftmost two black vertices (i*1*, i*2) in

e p p p u p p p u u e p p p u e e p p p e

1 *i*1 *i*2 *i*3 *ik* *n*

*o*

e p p p p p p u p p p u e p p p u e e p p p e

1 *i*2*− i*1 *i*3 *ik* *n*

*Let Dm* *= (im− im−1+ im−2− ... ± i*1*)(im+1− im+ im−1− im−2* *+ ... ∓ i*1) where

*0 < m < k. By Proposition 2.1(a), u ∼ (i*2 *− i*1*, i*3*, ..., ik) ∈ Vk−1* *needs at most D*1

steps.

Continuing the same way,

*(i*2*− i*1*, i*3*, ..., ik) ∼ (i*3 *− i*2 *+ i*1*, i*4*, ..., ik) ∈ Vk−2* *needs at most D*2 steps.

...

*(ik−1− ik−2+ ... ± i*1*, ik) ∼ (ik− ik−1+ ... ∓ i*1*) ∈ V*1 *needs at most Dk−1* steps.

*Then D*1 *≤ i*2*· i*2 *= i*22 *≤ n*2*, D*2 *≤ i*3*· i*3 *= i*32 *≤ n*2*, ... , Dk−1* *≤ ik· ik= ik*2 *≤ n*2,

*and thus d(u, r) = D*1 *+ D*2 *+ ... + Dk−1* *≤ n*2 *+ n*2 *+ ... + n*2 *= (k − 1)n*2, where

*r = (ik− ik−1+ ... ∓ i*1*) ∈ V*1.

*Case 2 If i*1*− 1 > n − ik, and just move the rightmost two black vertices (ik−1, ik*)

in the right direction:

e p p p e e u p p p e u u p p p u p p p e

1 *i*1 *ik−2ik−1* *ik* *n*

*o*

e p p p e e u p p p e u p p p u p p p p p p e

1 *i*1 *ik−2* *n + ik−1− ik+ 1 n*

*By Proposition 2.1(b), u ∼ (i*1*, ..., ik−2, n + ik−1− ik+ 1) ∈ Vk−1* *needs at most R*1

Continuously,

*(i*1*, ..., ik−2, n + ik−1− ik+ 1) ∼ (i*1*, ..., ik−3, ik− ik−1+ ik−2) ∈ Vk−2* *needs at most R*2

*steps where R*2 *= (ik− ik−1)(n + ik−1− ik+ 1 − ik−2*).

...

*(i*1*, ik− ik−1+ ik−2− ... ± i*2*) ∼ (n − ik+ ik−1− ... ∓ i*2*+ i*1*+ 1) ∈ V*1 needs at most

*Rk−1* *steps where Rk−1* *= (n − ik+ ik−1− ... ∓ i*2*+ i*1*+ 1)(ik− ik−1+ ik−2− ... ± i*1).

*Hence R*1 *≤ n · n = n*2*, R*2 *≤ n · n = n*2*, ... , Rk−1* *≤ n · n = n*2.

*Then d(u, t) = R*1 *+ R*2 *+ ... + Rk−1* *≤ n*2 *+ n*2 *+ ... + n*2 *= (k − 1)n*2*, where t =*

*(n − ik+ ik−1− ... ∓ i*2*+ i*1*+ 1) ∈ V*1.

### 3

### Type B

*A Dynkin diagram of type Bn* is given by

h h p p p h* _{=⇒}*h

1 2 *n − 1 n* _{(3.1)}

*A Vogan diagram of Bn* is labelled as (3.1) with at least one black vertex, denoted

*by V (Bn). Let Vk(Bn) ⊂ V (Bn) denote the Vogan diagram of Bn* *with k black*

*vertices. From (1.2), it also extends to the distance between Vk(Bn) and V*1*(Bn*) by

*d(Vk(Bn), V*1*(Bn)) = min{d(v, w); v ∈ Vk(Bn), w ∈ V*1*(Bn)}* (3.2)

*or just write dk(V*1*(Bn)). Note that if n is painted, then Fn* cannot reverse the color

*of n − 1, or we can write Fn(w) = w, here w ∈ V (Bn*)

*In this section, we discuss the distance between Vk(Bn) and V*1*(Bn*). We will

provide an upper bound for it.

*Proposition 3.1 Let u = (i, i + k) ∈ V (Bn), then u ∼ v needs at most ki steps*

*where v = (k).*

*P roof : Suppose that we want to obtain a Vogan diagram of Bn*with one black vertex

*from u immediately. Then we should move the painted vertices in the left direction,*
*or the vertex n will be painted and it does not make sense that we mentioned before.*

e p p p e u e p p p e e e p p p u p p p e* _{⇒}*e
1

*i*

*i + k*

*n*

*Fi*

*↓*e p p p u u u p p p e e e p p p u p p p e

*e 1*

_{⇒}*i − 1*

*i*

*i + 1*

*i + k*

*n*

*Fi+1*

*↓*e p p p u e u u p p p e e p p p u p p p e

*e 1*

_{⇒}*i − 1 i + 1*

*i + 2*

*i + k*

*n*

*F*

*↓*

...
*↓ Fi+k−2*
e p p p u e e e p p p p p p e u u u p p p e* _{⇒}*e
1

*i − 1*

*←*

*i + k − 2*

*→ i + k − 1*

*i + k*

*n*

*Fi+k−1*

*↓*e p p p u e e e p p p p p p e e u e p p p e

*e 1*

_{⇒}*i − 1*

*i + k − 1*

*n*

*The graph says that u = (i, i+k) −→ (i−1, i, i+1, i+k) −→ (i−1, i+1, i+2, i+k) −→*

*· · · −→ (i − 1, i + k − 2, i + k − 1, i + k) −→ (i − 1, i + k − 1) and it needs at most*

*[(i + k − 1) − i] + 1 = k steps.*
Continuously, we also can get that

*(i − 1, i + k − 1) ∼ (i − 2, i + k − 2) needs at most k steps.*
*(i − 2, i + k − 2) ∼ (i − 3, i + k − 3) needs at most k steps.*

...

*(2, 2 + k) ∼ (1, 1 + k) needs at most k steps.*

*Finally, (1, 1 + k) −→ (1, 2, 1 + k) −→ (2, 3, 1 + k) −→ · · · −→ (k − 1, k, k + 1) −→ (k)*
*and it needs at least k steps.*

*Thus, there are at most k + [(i − 1) − 1]k + k = ki steps to make (i, i + k) ∼ (k);*
*in other words, d(u, v) ≤ ki where v = (k).*

Therefore, we can conclude that:

*Proposition 3.2 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(Bn), then dk(V*1*(Bn)) ≤ (k − 1)n*2*.*

*P roof : By Proposition 3.1, there are at most i(j − i) steps to make (i, j) reduced to*

*(j − i).*

*Let u = (i*1*, i*2*, ..., ik) ∈ Vk*. Similar to Case 1 of Proposition 2.2, we move vertices

*and i*2. Repeat this process of moving pairs of leftmost black vertices towards the

*left. Finally, we prove that dk(V*1*(Bn)) ≤ (k − 1)n*2. Throughout these steps, vertex

*n remains unpainted.*

### 4

### Type C

*The Vogan diagram of Cn* *is similar to Bn*except for the direction of arrow on the

*double-edge, it is indicated by (4.1) and denoted by V (Cn*).

h h p p p h* _{⇐=}*h

1 2 *n − 1 n* _{(4.1)}

By Borel-Siebenthal Theorem, every Vogan diagram is equivalent to one with a
*black vertex. If a Vogan diagram belongs to V (Cn) and its vertex n is painted, then*

*this diagram is equivalent to (n). For example, (2, 4) ∈ V*4*(Cn) and then (2, 4) →*

*(2, 3, 4) → (3, 4) → (4).*

*Before talking about the distance between Vk(Cn) and V*1*(Cn*), we consider the

*distance between V*2*(Cn) and V*1*(Cn*) first.

*Proposition 4.1 Let u = (i, j) ∈ V (Cn).*

*(a) If j 6= n (i.e. n is unpainted), then*

*(i) d(u, v) ≤ i(j − i) if i − 1 ≤ (n − 1) − j, where v = (j − i).*

*(ii) d(u, w) ≤ (n − j)(j − i) if i − 1 ≥ (n − 1) − j, where w = (n − j + i).*
*(b) If j = n (i.e. n is painted), then d(u, s) ≤* *(n−i)(n−i+1)*_{2} *where s = (n).*

*P roof :*

*(a) If n is unpainted and we cannot use any method to make it become painted,*
*then we can ignore the existence of n or just regard this Vogan diagram as V (An−1*).

Recall the result of Proposition 2.1, then the proof of (i) and (ii) can be completed right away.

*(b) If j = n, then u = (i, n).*

*Since n is black, there are no ways to let it change color. Now, we want to move the*
*first black vertex closer to n(in other words, move it in the right direction), or they*
are too far to get the major distance.

e p p p e u e e p p p e e e* _{⇐}*u
1

*i*

*n*

*Fn*

*↓*e p p p e u e e p p p e e u

*u 1*

_{⇐}*i*

*n − 1*

*n*

*Fn−1*

*↓*e p p p e u e e p p p e u u

*u 1*

_{⇐}*i*

*n − 2*

*n − 1*

*n*

*Fn−2*

*↓*e p p p e u e e p p p u u e

*u 1*

_{⇐}*i*

*n − 3*

*n − 2*

*n*

*Fn−3*

*↓*...

*Fi+2*

*↓*e p p p e u u u p p p e e e

*u 1*

_{⇐}*i*

*i + 1*

*i + 2*

*n*

*↓ Fi+1*e p p p e e u e p p p e e e

*u 1*

_{⇐}*i + 1*

*n*

*The graph says that u = (i, n) −→ (i, n − 1, n) −→ (i, n − 2, n − 1, n) −→ (i, n − 3,*

*n−2, n) −→ · · · −→ (i, i+1, i+2, n) −→ (i+1, n) and it needs at most n−(i+1)+1*

*= n − i steps.*

By the same way, we can get that

*(i + 1, n) ∼ (i + 2, n) needs at most n − i − 1 steps.*
*(i + 2, n) ∼ (i + 3, n) needs at most n − i − 2 steps.*

...

*(n − 2, n) ∼ (n − 1, n) needs at most 2 steps.*
*(n − 1, n) −→ (n) needs 1 steps.*

=
*n−i*
X
*k=1*
*k =* *(n − i)(n − i − 1)*
2 .
Therefore, we can conclude that:

*Proposition 4.2 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(Cn).*
*(a) If ik* *6= n, then dk(V*1*(Cn)) ≤ (k − 1)(n − 1)*2*.*
*(b) If ik* *= n, then dk(V*1*(Cn)) ≤* *(k−1)n*
2
2 *.*
*P roof :*

*(a) If ik6= n, then u ∈ V (An−1*). Now, there are two cases to discuss:

*i*1*− 1 ≤ (n − 1) − ik*

*i*1*− 1 > (n − 1) − ik*

*By Proposition 2.2, the result of (a) can be proved.*

*(b) First, we want to reduce the number of painted vertices step by step. With*
*the same idea as Proposition 4.1(2), we move the first k − 1 black vertices closer to*

*n. Therefore, we just move ik−1* *rightwards and make ik−1* *and n reduce to one black*

vertex.
e p p p u e p p p u e u p p p e u p p p e* _{⇐}*u
1

*i*1

*i*2

*i*3

*ik−1*

*n*

*o*e p p p u e p p p u e u p p p e e p p p e

*u 1*

_{⇐}*i*1

*i*2

*i*3

*n*

*Let Sp* = 1_{2}*(n − ik−p)(n − ik−p+ 1) where 0 < p < k. By Proposition 4.1(b),*

*u = (i*1*, i*2*, · · · , ik−1, n) ∼ (i*1*, i*2*, ik−2, · · · , n) ∈ Vk−1(Cn) needs at most S*1 steps.

Continuously,

*(i*1*, i*2*, · · · , ik−2, n) ∼ (i*1*, i*2*, ik−3, · · · , n) ∈ Vk−2(Cn) needs at most S*2 steps.

...

*(i*1*, n) ∼ (n) ∈ V*1*(Cn) needs at most Sk−1* steps.

*Then S*1 *≤* 1_{2}*n · n =* *n*
2
2 *, S*2 *≤* *n*
2
2 *, S*3 *≤* *n*
2
2 *, ... , Sk−1* *≤* *n*
2
2 *and thus dk(V*1*(Cn*)) =
*S*1*+ S*2*+ · · · + Sk−1* *≤* *n*
2
2 *(k − 1).*

### 5

### Type D

*A Vogan diagram of Dn* *is a Dynkin diagram of Dn* *with a diagram involution θ,*

*denote it by V (Dn, θ)(see (5.1)). Besides, θ can be trivial and we also call it a Vogan*

*diagram of Dn, denote it by V (Dn) (see (5.2)). However the vertices n − 1 and n*

*fixed by θ can be painted.*

h h p p p h¡
¡
@
@
h
h
*l θ*
1 2 *n − 2*
*n − 1*
*n*
(5.1)
h h p p p h¡
¡
@
@
h
h
1 2 *n − 2*
*n − 1*
*n*
(5.2)
About (5.1) and (5.2), we give two examples to compare the difference between

*V (Dn, θ) and V (Dn*) if we use the same process.

*Example :*
(1)
e e u e u¡¡
@@
e
e
*l θ −→F*5 e e u u u¡¡
@@
e
e
*l θ −→F*4 e e e u e¡¡
@@
e
e
*l θ*

*( i.e. (3, 5) ∈ V (D*7*, θ) and (3, 5) ∼ (4) needs 2 steps. )*

(2)
e e u e u¡¡
@@
e
e
*−→F*5 e e u u u¡¡
@@
u
u
*−→F*4 e e e u e¡¡
@_{@}
u
u
*−→F*6 _{e} _{e} _{e} _{u} _{u¡}¡
@@
u
u
*−→F*5 e e e e u¡¡
@@
e
e

*( i.e. (3, 5) ∈ V (D*7*) and (3, 5) ∼ (5) needs 4 steps. )*

*Recall that Vk(Dn) means there are k black vertices in Vk(Dn*). From (1.2), it also

*extends to the distance between Vk(Dn) and V*1*(Dn*) by

*d(Vk(Dn), V*1*(Dn)) = min{d(v, w); v ∈ Vk(Dn), w ∈ V*1*(Dn)}* (5.3)

*or just write dk(V*1*(Dn*)). In this section, we provide an upper bound for the distance

*between Vk(Dn) and V*1*(Dn). Now, we discuss V (Dn, θ) as follows.*

*Proposition 5.1 Let u = (i, j) ∈ V (Dn, θ).*

*(a) If i − 1 ≤ (n − 2) − j, then d(u, v) ≤ i(j − i) where v = (j − i).*

*(b) If i − 1 > (n − 2) − j, then d(u, w) ≤ (n − j − 1)(j − i) where w = (n − j + i − 1).*

*P roof : Obviously, we can regard V (Dn, θ) as V (An−2) since n − 1 and n have no*

*colors. Then u = (i, j) ∈ V (An−2) and by Proposition 2.1, the results of (a) and (b)*

here will be proved.

Therefore, we can conclude that:

*Proposition 5.2 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(Dn, θ), then dk(V*1*(Dn, θ)) ≤ (k − 1) ·*

*(n − 2)*2_{.}

*P roof : Similarly, V (Dn, θ) is equivalent to V (An−2*). Extending the result of

*Propo-sition 2.2, we obtain that dk(V*1*(Dn, θ)) ≤ (k − 1)(n − 2)*2 right away.

*Consequently, we can get the result of Theorem 1 (e) right away. Continuously,*
*we try to find the upper bound for d(Vk(Dn), V*1*(Dn)) and then discuss dk(V*1*(Dn*)).

*Proposition 5.3 Let u = (i, j) ∈ V (Dn).*

*(a) If j /∈ {n − 1, n}, then d(u, v) ≤ i(j − i) where v = (j − i).*

*(b) If i /∈ {n−1, n} and j ∈ {n−1, n}, then d(u, w) ≤* *(n−i)(n−i−1)*_{2} *where w = (n−1)*
*or (n).*

*P roof :*

*(a) Suppose that j /∈ {n − 1, n}. Then i, j ∈ {1, 2, ..., n − 2}. Now, we want to*

*reduce u to be a Vogan diagram with one painted vertex. If we move these two black*
*vertices toward right, then vertices n − 1 and n will become painted and the path to*
*a Vogan diagram of V*1*(Dn*) must be more complicated. Hence we move black vertices

toward left.
e e p p p e u e p p p e e u p p p e¡¡
@@
e
e
*i* *j*
*↓ Fi*
e e p p p u u u p p p e e u p p p e¡¡
@@
e
e
*i − 1*
*i*
*i + 1*
*j*
*↓ Fi+1*
...
*↓ Fj−2*
e e p p p u e e p p p u u u p p p e¡¡
@@
e
e
*j − 2*
*i − 1* *j*
*↓ Fj−1*
e e p p p u e e p p p e u e p p p e¡¡
@@
e
e
*i − 1* *j − 1*

*It means that u = (i, j) ∼ (i − 1, j − 1) needs at most (j − i) steps.*
Continuously,

*(i − 1, j − 1) ∼ (i − 2, j − 2) needs at most (j − i) steps.*
*(i − 2, j − 2) ∼ (i − 3, j − 3) needs at most (j − i) steps.*

...

*(1, j − i + 1) ∼ (j − i) needs at most (j − i) steps.*

*So, there are at most i(j − i) steps to make (i, j) ∼ (j − i); in other words,*

*(b) Suppose that i /∈ {n−1, n} and j /∈ {n−1, n}. Then u = (i, n−1) or u = (i, n),*

*0 < i < n−1. We only talk about u = (i, n−1) because (i, n) is equivalent to (i, n−1),*
and the conclusions are the same.

*Now, we want to move these two black vertices closer and u will be reduced to a*
*Vogan diagram of V*1*(Dn*) as soon as possible.

e e p p p u e e p p p e e e¡¡
@@
u
e
*i*
*j = n − 1*
*↓ Fn−1*
e e p p p u e e p p p e e u¡¡
@@
u
e
*i* *n − 2*
*j = n − 1*
*↓ Fn−2*
e e p p p u e e p p p e u u¡¡
@@
e
u
*i* *n − 3* _{n}*↓ Fn−3*
e e p p p u e e p p p u u e¡¡
@@
e
u
*i* *n − 4*
*n*
*↓ Fn−4*
...
*↓ Fi+2*
e e p p p u u u p p p e e e¡¡
@@
e
u
*i*
*i + 1*
*i + 2*
*n*
*↓ Fi+1*
e e p p p e u e p p p e e e¡¡
@@
e
u
*i + 1*
*n*

*We get (i, n − 1) ∼ (i + 1, n) needs at most (n − 1) − (i − 1) + 1 = n − i − 1 steps.*
By the same way, we also get

*(i + 1, n) ∼ (i + 2, n − 1) needs at most n − i − 2 steps.*
*(i + 2, n − 1) ∼ (i + 3, n) needs at most n − i − 3 steps.*

...

*(n − 2, n − 1) ∼ (n) or (n − 2, n) ∼ (n − 1) needs at most 1 steps.*

*Thus, d(u, w) ≤ 1 + 2 + · · · + (n − i − 1) =* *(n−i)(n−i−1)*_{2} *where w = (n) or (n − 1).*
*(c) We have known that vertices n − 1 and n are painted. The only way to let u*
be a diagram with one painted vertex is moving them in the left direction.

e e p p p e e p p p e e¡¡
@@
u
u
*n − 1*
*n*
*↓Fn* *or Fn−1*
e e p p p e e p p p e u¡¡
@_{@}
u
u
*n − 2*
*n − 1*
*n*
*↓Fn−2*
e e p p p e e p p p u u¡¡
@@
e
e
*n − 2*
*n − 1*
*n*
*↓Fn−3*
...
*↓F*2
u u p p p e e p p p e e¡¡
@@
e
e
1 2
*↓F*1
u e p p p e e p p p e e¡¡
@@
e
e
1

*From the above graph, it tells us that (n − 1, n) ∼ (1) needs at most (n − 1) − 1 + 1*
*steps, i.e. d(u, p) ≤ n − 1 where p = (1).*

Therefore, we can conclude that:
*Proposition 5.4 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(Dn).*
*(a) If n − 1, n /∈ u, then dk(V*1*(Dn)) ≤ (k − 1)(n − 2)*2*.*
*(b) If n − 1 ∈ u or n ∈ u, then dk(V*1*(Dn)) ≤* *(k−1)(n−1)*
2
2 *.*
*(c) If n − 1, n ∈ u, then dk(V*1*(Dn)) ≤ (k − 3)(n − 2)*2*+ (n − 2).*
*P roof :*

*(a) Suppose that 1 ≤ i*1*, i*2*, · · · , ik≤ n−2 (i.e. vertices n−1 and n are unpainted).*

Looking back the method of Proposition 5.3(a), we just try to move the leftmost two black vertices toward left and then become one black vertex. Step by step, we can finally get only one black vertex in this Vogan diagram.

e p p p u e p p p u e u p p p u e p p p u e¡¡
@@
e
e
*i*1 *i*2 *i*3 *ik−1* *ik*
*o*
e p p p e p p p u p p p u p p p u e p p p u e¡¡
@@
e
e
*i*2*− i*1 *i*3 *ik−1* *ik*

*Let Rm* *= (im− im−1+ im−2− ... ± i*1*)(im+1* *− im+ im−1− im−2+ ... ∓ i*1) where

*0 < m < k. By Proposition 5.3(a), u ∼ (i*2 *− i*1*, i*3*, ..., ik) ∈ Vk−1(Dn*) needs at most

*R*1 steps. Continuing the same way,

*(i*2*− i*1*, i*3*, ..., ik) ∼ (i*3 *− i*2 *+ i*1*, i*4*, ..., ik) ∈ Vk−2(Dn) needs at most R*2 steps.

...

*(ik−1− ik−2+ ... ± i*1*, ik) ∼ (ik− ik−1+ ... ∓ i*1*) ∈ V*1*(Dn) needs at most Rk−1* steps.

*Then R*1 *≤ i*2*· i*2 *= i*22 *≤ (n − 2)*2*, R*2 *≤ i*3*· i*3 *= i*32 *≤ (n − 2)*2*,· · ·, Rk−1* *≤ ik· ik*=

*ik*2 *≤ (n − 2)*2.

*Thus d(u, r) = R*1*+R*2*+...+Rk−1* *≤ (n−2)*2*+(n−2)*2*+...+(n−2)*2 *= (k−1)(n−2)*2,

*(b) If n − 1 ∈ u or n ∈ u, then one of vertices n − 1 and n − 1 is painted (i.e.*

*ik* *= n − 1 or ik= n, but ik−1* *6= n − 1).*

*Similar to the process of Proposition 5.3(b), we try to move the black vertex ik−1*

towards the right. Finally, these two vertex will be reduced to a black vertex.
*Here we just talk about ik* *= n − 1 because its result is the same as ik* *= n.*

e p p p u e p p p u e u p p p u u p p p e e¡¡
@@
u
e
*i*1 *i*2 *i*3 *ik−2ik−1*
*ik* *= n − 1*
*o*
e p p p u e p p p u e u p p p u e p p p e e¡¡
@@
u
e
*i*1 *i*2 *i*3 *ik−2*
*n − 1*

*Note that the rightmost black vertex could be n − 1 or n, but it does not affect*
the result. Let

*Sp* = 1

2*(n − ik−p)(n − ik−p− 1), 1 ≤ p ≤ k − 1*
. By Proposition 5.3(b),

*u ∼ (i*1*, i*2*, ..., ik−2, n − 1) needs at most S*1 steps. With the same method,

*(i*1*, i*2*, ..., ik−2, n − 1) ∼ (i*1*, i*2*, ..., ik−3, n − 1) needs at most S*2 steps.

*(i*1*, i*2*, ..., ik−3, n − 1) ∼ (i*1*, i*2*, ..., ik−4, n − 1) needs at most S*3 steps.

...

*(i*1*, n − 1) ∼ (n − 1) needs at most Sk−1* steps.

*We find that S*1 *≤* 1_{2}*(n−2)(n−1), S*2 *≤* 1_{2}*(n−2)(n−1), · · · , Sk−1* *≤* 1_{2}*(n−2)(n−1).*

*Thus d(u, w) = S*1*+ S*2*+ · · · + Sk−1* *≤* 1_{2}*(k − 1)(n − 2)(n − 1), where w = (n − 1) or*

*(n) and this proof is completed.*

*(c) Let ik−1* *= n − 1 and ik* *= n, i.e. u = (i*1*, i*2*, ..., ik−2, n − 1, n). Regardless of*

*vertices n − 1 and n, we hope to reduce the k − 2 black vertices in 1, · · · , n − 1 to one*
*black vertex. Then adding in vertices n − 1 and n, we finally make these three black*
vertices become only one black vertex.

Referring to the method of (a), we move the leftmost two painted vertices towards
the left.
e p p p u e p p p u e u p p p u u p p p e e¡¡
@@
u
u
*i*1 *i*2 *i*3 *ik−2*
*n − 1*
*n*
*o*
e p p p e e u p p p e u p p p u u p p p e e¡¡
@@
u
u
*i*2*− i*1 *i*3 *ik−2*
*n − 1*
*n*
*o*
...
*o*
e p p p e e e p p p e e p p p u e p p p e e¡¡
@@
u
u
*n − 1*
*n*

*Let Ms* *= (is− is−1* *+ · · · ± i*1*)(is+1− is+ is−1− · · · ∓ i*1*) where 1 ≤ s ≤ k − 3.*

*By Proposition 5.3(a), u ∼ (i*2*− i*1*, i*3*, . . . , n − 1, n) needs at most M*1 steps. Keep on

the same way:

*(i*2*− i*1*, i*3*, . . . , n − 1, n) ∼ (i*3*− i*2*+ i*1*, i*4*, . . . , n − 1, n) needs at most M*2 steps.

*(i*3*− i*2 *+ i*1*, i*4*, . . . , n − 1, n) ∼ (i*4 *− i*3 *+ i*2 *− i*1*, i*5*, . . . , n − 1, n) needs at most M*3

steps.

...

*(ik−3− ik−4+ ... ± i*1*, ik−2, n − 1, n) ∼ (ik−2− ik−3+ ... ∓ i*1*, n − 1, n) needs at most*

*Mk−3* steps.

*Let a = ik−2* *− ik−3* *+ ... ∓ i*1 *and the final step is to find vertex i such that*

*(a, n − 1, n) ∼ (i):*

*(a, n − 1, n) −→ (a, n − 2, n − 1, n) −→ (a, n − 3, n − 2) −→ (a, n − 4, n − 3) −→*

*· · · −→ (a, a + 1, a + 2) −→ (a + 1) needs at most (n − a − 1) steps and i = a + 1.*

*We have known that 1 ≤ a ≤ n − 2, then 1 ≤ n − a − 1 ≤ n − 2.*

*≤ i*1*· i*2*+ i*2 *· i*3 *+ · · · + ik−3· ik−2+ (n − a − 1)*

*≤ (n − 2)*2* _{+ (n − 2)}*2

_{+ · · · + (n − 2)}*= (k − 3)(n − 2)*2_{+ (n − 2).}

### 6

### Type E

_{6}

*After observing An, Bn, Cnand Dn, we talk about other different type: E*6*, E*7*, E*8*, F*4

*and G*2.
e e e e e
e
E6
e e e e e e
e
E7
e e e e e e e
e
E8
e e* _{=⇒}*e e
F4
e

*e G2 (6.1)*

_{=⇒}*This section will deal with the upper bound for the distance between Vk(E*6) and

*V*1*(E*6*). Others are discussed later. Besides (6.1), there is another type of E*6, unfixed

*by θ, denoted by V (E*6*, θ).*

e e e e e e

*θ*

*↑* *↑*

e e e e e
e
1 2 3 4 5
0
e e e e e
e
1 2 3 4 5
0
*θ*
*↑* *↑*

*We give an example about V (E*6*) and V (E*6*, θ) on how to reduce some black*

vertices to only one black vertex.

*Example :*
*V (E*6): u e u e u
e
*−→F*3 u u u u u
u
*−→F*2 e u e u u
u
*−→F*4
e u u u e
u
*−→F*3 e e u e e
e
*i.e. (1,3,5)−→(0,1,2,3,4,5)−→(0,2,4,5)−→(0,2,3,4)−→(3)*
*V (E*6*, θ):* e e u e e
u
*↑* *↑*
*θ*
*−→F*0 e e e e e
u
*↑* *↑*
*θ*
*i.e. (0,3)−→(0)*

*Indeed, there are two vertices in V (E*6*, θ) can be drawn color, vertex 0 and vertex*

*3, just like the vertices n and n − 1 in V (Dn, θ).*

Recall that

*dk(V*1*(E*6*)) = d(Vk(E*6*), d*1*(E*6*)) = min{d(u, v); u ∈ Vk(E*6*), v ∈ V*1*(E*6*)}*

*Proposition 6.1 dk(V*1*(E*6*)) ≤ 5(5k − 4) where 1 ≤ k ≤ 6.*

*P roof : Separate E*6 *into A*5 and a vertex 0. Regardless of vertex 0, we reduce other

*black vertices to be one black vertex in V (A*5). There are 4 cases to discuss.

Case 1 Consider vertex 0 is white and all black vertices are in vertex 1 to 3 or vertex 3 to 5. Then vertex 0 will not become painted. By Proposition 2.2, we

*can get the distance between Vk(E*6*) to V*1*(E*6*) directly. dk(V*1*(E*6*)) = dk(V*1*(A*5*)) ≤*

52_{(k − 1) = 25(k − 1).}

Case 2 Consider vertex 0 is black and all black vertices are in vertex 1 to 3
or vertex 3 to 5. Then vertex 0 will not become unpainted. We arrange the black
*vertices in V (A*5) to be singly white and finally deal with vertex 0. We have to discuss

the following cases.

*(i) u ∼ (0, 1) and (0, 1) −→ (0, 1, 2) −→ (0, 2, 3) −→ (3, 4) −→ (4, 5) −→ (5). By*
*Proposition 2.2, we have known that dk(V*1*(A*5*)) ≤ 5*2*(k − 1) = 25(k − 1). Then*

*d(u, (5)) ≤ 25(k − 1) + 5 = 25k − 20.*

*(ii) u ∼ (0, 2) and (0, 2) −→ (0, 2, 3) −→ (3, 4) −→ (4, 5) −→ (5).*
*Then d(u, (5)) ≤ 25(k − 1) + 4 = 25k − 21.*

*(iii) u ∼ (0, 3) and (0, 3) −→ (0). Then d(u, (5)) ≤ 25(k − 1) + 1 = 25k − 24.*

*We don’t handle with (0, 4) and (0, 5) because they are equivalent to (0, 2) and*
*(0, 1). Thus, dk(V*1*(E*6*)) ≤ max{25k − 20, 25k − 21, 25k − 24} = 25k − 20 where*

*1 ≤ k ≤ 6.*

Case 3 Consider vertex 0 is white. Some black vertices are in vertex 1 to 3
and others are in vertex 3 to 5. Then the process of becoming singly painted will
affect the color of vertex 0. We let vertex 0 be black and use the same method as
*Case 2. Similarly, there are 3 cases to discuss and thus dk(V*1*(E*6*)) ≤ 25k − 20 where*

*1 ≤ k ≤ 5.*

Case 4 Consider vertex 0 is black. Some black vertices are in vertex 1 to 3 and
others are in vertex 3 to 5. The color of vertex 0 will be reversed in the process of
*reducing to singly painted vertex. We let vertex 0 be white and regard V (E*7) as

*V (A*6*). Thus, dk(V*1*(E*6*)) = dk(V*1*(A*5*)) ≤ 25(k − 1) = 25k − 25 where 1 ≤ k ≤ 6.*

*Arrange above 4 cases, we get that dk(V*1*(E*6*)) ≤ 25k − 20 = 5(5k − 4) where*

*Proposition 6.2 dk(V*1*(E*6*, θ)) ≤ 1 where k = 2.*

*P roof : Since vertices 1, 2, 4 and 5 are unfixed by θ, we do not have to paint any*

color on them. In other words, only vertices 0 and 3 can have color. Hence, the only
*one case is like the above example, (0,3)−→(0), needs only one step. Therefore, we*
get the proof of Theorem 2(b) immediately.

### 7

### Type E

_{7}

*By (6.1), we also label 0,1,...,7 to the Vogan diagram of E*7.

e e e e e e e

1 2 3 4 5 6 0

*The process of making some black vertices to be singly painted is similar to V (E*6).

*Now, we discuss how many steps that Vk(E*7*) goes to V*1*(E*7) at most.

*Proposition 7.1 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(E*7*), dk(V*1*(E*7*)) ≤ 12(3k − 2) where*

*1 ≤ k ≤ 7.*

*P roof : Separate E*7 *into A*6 and a vertex 0. We try to discuss the distance between

*k painted vertices and only one painted vertex by the color of vertex 0.*

Case 1 Suppose that vertex 0 is white and all black parts are in vertices 1 to 3
or vertices 3 to 6. Then the process of reducing to only one black vertex cannot affect
*the color of vertex 0. Thus we just regard V (E*7*) as V (A*6) and by Proposition 2.2,

*dk(V*1*(E*7*)) = dk(V*1*(A*6*)) ≤ 6*2*(k − 1) = 36k − 36.*

Case 2 Suppose that vertex 0 is black and other black parts are in vertices 1 to
3 or vertices 3 to 6. Then the process of reducing to only one black vertex cannot
*affect the color of vertex 0. We make these painted vertices in V (A*6) become one

painted and then act on vertex 0. There are 6 cases to be discussed.

*(i) u ∼ (0, 1) and (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6).*
*By Proposition 2.2, we have known that dk(V*1*(A*6*)) ≤ 6*2*(k − 1) = 36(k − 1).*

*Then d(u, (6)) ≤ 36(k − 1) + 6 = 36k − 30.*

*(ii) u ∼ (0, 2) and (0, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6).*
*Then d(u, (6)) ≤ 36(k − 1) + 5 = 36k − 31.*

*(iii) u ∼ (0, 3) and (0, 3) → (0).*

*(iv) u ∼ (0, 4) and (0, 4) → (0, 3, 4) → (2, 3) → (1, 2) → (1).*
*Then d(u, (1)) ≤ 36(k − 1) + 4 = 36k − 32.*
*(v) u ∼ (0, 5) and (0, 5) → (0, 3, 5) → (2, 3, 4, 5) → (2, 4) → (1, 2, 3, 4) → (0, 1, 3) →*
*(0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6).*
*Then d(u, (6)) ≤ 36(k − 1) + 12 = 36k − 24.*
*(vi) u ∼ (0, 6) and (0, 6) → (0, 5, 6) → (0, 4, 5) → (0, 3, 4) → (2, 3) → (1, 2) → (1).*
*Then d(u, (1)) ≤ 36(k − 1) + 6 = 36k − 30.*
*Thus, dk(V*1*(E*7*)) ≤ max{36k−30, 36k−31, 36k−35, 36k−32, 36k−24, 36k−30} =*
*36k − 24 where 1 ≤ k ≤ 7.*

Case 3 Suppose that vertex 0 is white and some black vertices are in vertex 1 to 3 and the other black in vertex 3 to 6. The color of vertex 0 will be reversed in the process of reducing to singly painted vertex. We let vertex 0 be black and use the same method as Case 2. Similarly, there are 6 cases to discuss and thus

*dk(V*1*(E*7*)) ≤ 36k − 24 where 1 ≤ k ≤ 6.*

Case 4 Suppose that vertex 0 is black and some black vertices are in vertex 1 to
3 and others are in vertex 3 to 6. The color of vertex 0 will be reversed in the process
*of reducing to singly painted vertex. We let vertex 0 be white and look V (E*7) as

*V (A*6*). Thus, dk(V*1*(E*7*)) = dk(V*1*(A*6*)) ≤ 36(k − 1).*

*Combining with Case 1 to Case 4, we can get that dk(V*1*(E*7*)) ≤ 36k − 24 =*

### 8

### Type E

_{8}

*From the diagram of (6.1), we have known that what the Vogan diagram of E*8

*looks like. Similarly, we give 8 numbers on E*8.

e e e e e e e e

1 2 3 4 5 6 7 0

*This section also talk about what is the maximal distance between Vk(E*8) and

*V*1*(E*8).

*Proposition 8.1 Given u = (i*1*, i*2*, ..., ik) ∈ Vk(E*8*), then dk(V*1*(E*8*)) ≤ 5(5k − 4)*

*where 1 ≤ k ≤ 8.*

*P roof : Use the same way as Proposition 6.1 and Proposition 7.1, we also have to*

talk about 4 cases. As above, we just calculate Case 2 and Case 3 since they need more steps.

*I. Suppose that vertex 0 is black and other black parts are in vertices 1 to 3 or*
vertices 3 to 7. Then the process of reducing to only one black vertex cannot affect
*the color of vertex 0. We make these painted vertices in V (A*7) become one painted

and then act on vertex 0. There are 7 cases to be discussed.

*(i) u ∼ (0, 1) and (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) →*
(7).

*By Proposition 2.2, we have known that dk(V*1*(A*7*)) ≤ 7*2*(k − 1) = 49(k − 1).*

*Then d(u, (7)) ≤ 49(k − 1) + 7 = 49k − 42.*
*(ii) u ∼ (0, 2) and (0, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) → (7).*
*Then d(u, (7)) ≤ 49(k − 1) + 6 = 49k − 43.*
*(iii) u ∼ (0, 3) and (0, 3) → (0).*
*Then d(u, (0)) ≤ 49(k − 1) + 1 = 49k − 48.*
*(iv) u ∼ (0, 4) and (0, 4) → (0, 3, 4) → (2, 3) → (1, 2) → (1).*

*Then d(u, (1)) ≤ 49(k − 1) + 4 = 49k − 45.*
*(v) u ∼ (0, 5) and (0, 5) → (0, 3, 5) → (2, 3, 4, 5) → (2, 4) → (1, 2, 3, 4) → (0, 1, 3) →*
*(0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) → (7).*
*Then d(u, (7)) ≤ 49(k − 1) + 13 = 49k − 36.*
*(vi) u ∼ (0, 6) and (0, 6) → (0, 3, 6) → (2, 3, 4, 6) → (2, 4, 5, 6) → (2, 5) → (1, 2, 3, 5)*
*→ (0, 1, 3, 4, 5) → (0, 1, 4) → (0, 1, 3, 4) → (1, 2, 3) → (2).*
*Then d(u, (1)) ≤ 49(k − 1) + 10 = 49k − 39.*
*(vii) u ∼ (0, 7) and (0, 7) → (0, 6, 7) → (0, 5, 6) → (0, 4, 5) → (0, 3, 4) → (2, 3) →*
*(1, 2) → (1).*
*Then d(u, (1)) ≤ 49(k − 1) + 7 = 49k − 42.*
*Thus, dk(V*1*(E*8*)) ≤ max{49k − 42, 49k − 43, 49k − 48, 49k − 45, 49k − 36, 49k −*
*39, 49k − 42} = 49k − 36 where 1 ≤ k ≤ 8.*

*II. Suppose that vertex 0 is white and some black vertices are in vertex 1 to 3 and*
the other black in vertex 3 to 7. The color of vertex 0 will be reversed in the process of
reducing to singly painted vertex. We let vertex 0 be black and use the same method
*as I. Similarly, there are 7 cases to discuss and thus dk(V*1*(E*8*)) ≤ 49k − 36 where*

*1 ≤ k ≤ 7.*

*Combining with I and II, we can get that dk(V*1*(E*8*)) ≤ 49k −36 where 1 ≤ k ≤ 8.*

### 9

### Type F

_{4}

### and G

_{2}

*Finally, we look the remaining Vogan diagram, F*4 *and G*2 and observe that the

*upper bound for d(Vk(F*4*), V*1*(F*4*)) (k = 2, 3, 4) and d(Vk(G*2*, V*1*(G*2)).

*Proposition 9.1 dk(V*1*(F*4*)) ≤ 4 where k = 2, 3, 4.*

*P roof : Since there are only 4 vertices in V (F*4), we just use diagram to find the

*distance between Vk(F*4*) and V*1*(F*4*), k = 2, 3, 4.*

u u* _{=⇒}*e e

*u e*

_{−→}*e e u e*

_{=⇒}*u e*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e u e*

_{=⇒}*e u*

_{=⇒}*u u*

_{−→}*e u*

_{=⇒}*u u*

_{−→}*u u*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e e u*

_{=⇒}*u e*

_{=⇒}*u u*

_{−→}*e e*

_{=⇒}*u e*

_{−→}*e e e u*

_{=⇒}*e u*

_{=⇒}*u u*

_{−→}*u u*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e e e*

_{=⇒}*u u*

_{=⇒}*e e*

_{−→}*e u u u*

_{=⇒}*u e*

_{=⇒}*e u*

_{−→}*e e u u*

_{=⇒}*e u*

_{=⇒}*u u*

_{−→}*u u*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e u e*

_{=⇒}*u u*

_{=⇒}*u u*

_{−→}*u u*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e e u*

_{=⇒}*u u*

_{=⇒}*e u*

_{−→}*u e*

_{=⇒}*u u*

_{−→}*e e*

_{=⇒}*u e*

_{−→}*e e u u*

_{=⇒}*u u*

_{=⇒}*u u*

_{−→}*u e*

_{=⇒}*e u*

_{−→}*e e*

_{=⇒}From the above process, we can find that it needs at most 4 steps to make a Vogan diagram with some black vertices reduce to only one black vertex.

*Proposition 9.2 dk(V*1*(G*2*)) ≤ 1 where k = 2.*

*P roof : V (G*2) is graphed as e*=⇒*eand there are at most 2 vertices to be painted.

*So, we just talk about the only one case, (1, 2).*

u* _{=⇒}*u

*u*

_{−→}*e*

_{=⇒}### References

[1] A. Borel and J. de Siebenthal, Les sous-groupes fermes de rang maximum des
*groupes de Lie clos, Comment. Math. Helv. 23 (1949), 200-221.*

*[2] J. E. Humphreys, Introduction to Lie Algebras and Representation Theory,*
Springer-Verlag, New York 1972.